Question

1. \\(\frac{a - 2}{a + 3} - 1 = \frac{3}{a + 2}\\)

Explanation:

Step1: Find common denominator

The common denominator of \(a + 3\), \(1\) (which can be written as \(\frac{a + 3}{a + 3}\)), and \(a + 2\) is \((a + 3)(a + 2)\). Multiply each term by \((a + 3)(a + 2)\) to eliminate the fractions:

$$ (a - 2)(a + 2) - (a + 3)(a + 2) = 3(a + 3) $$

Step2: Expand the expressions

Expand each product using the distributive property (FOIL method for binomials):

• For \((a - 2)(a + 2)\), we use the difference of squares formula \( (x - y)(x + y)=x^{2}-y^{2}\), so \((a - 2)(a + 2)=a^{2}-4\).
• For \((a + 3)(a + 2)\), we have \(a^{2}+2a + 3a+6=a^{2}+5a + 6\).
• For \(3(a + 3)\), we get \(3a+9\).

Substituting these back into the equation:

$$ a^{2}-4-(a^{2}+5a + 6)=3a + 9 $$

Step3: Simplify the left - hand side

Distribute the negative sign in front of \((a^{2}+5a + 6)\):

$$ a^{2}-4 - a^{2}-5a - 6=3a + 9 $$

Combine like terms. The \(a^{2}\) terms cancel out:

$$ -5a-10 = 3a + 9 $$

Step4: Solve for \(a\)

Add \(5a\) to both sides of the equation:

$$ -10=8a + 9 $$

Subtract \(9\) from both sides:

$$ -19 = 8a $$

Divide both sides by \(8\):

$$ a=-\frac{19}{8} $$

Step5: Check for extraneous solutions

We need to check if \(a = -\frac{19}{8}\) makes the original denominators zero. The denominators are \(a + 3=-\frac{19}{8}+3=-\frac{19}{8}+\frac{24}{8}=\frac{5}{8}
eq0\) and \(a + 2=-\frac{19}{8}+2=-\frac{19}{8}+\frac{16}{8}=-\frac{3}{8}
eq0\). So \(a = -\frac{19}{8}\) is a valid solution.

Answer:

\(a = -\frac{19}{8}\)